1. 3 masses hang as shown, they are connected by light strings and your basic frictionless pulley. (a) find…

1. 3 masses hang as shown, they are connected by light strings and your basic frictionless pulley. (a) find the acceleration of each mass and (b) the tensions in the 2 strings.\n2. a 20.0 kg cart with very low friction wheels sits on a table. a light string is attached to it and runs over a low friction pulley to a 0.0150 kg mass. what is the acceleration experienced by the cart?

1. 3 masses hang as shown, they are connected by light strings and your basic frictionless pulley. (a) find the acceleration of each mass and (b) the tensions in the 2 strings.\n2. a 20.0 kg cart with very low friction wheels sits on a table. a light string is attached to it and runs over a low friction pulley to a 0.0150 kg mass. what is the acceleration experienced by the cart?

Answer

Explanation:

Step1: Analyze forces on masses in first - problem

For the three - mass system, let the acceleration of the system be (a) and the tensions in the strings be (T_1) and (T_2). For (m_1 = 3.00\ kg), (m_1g-T_1=m_1a). For (m_2 = 4.00\ kg), (T_1 - m_2g+T_2=m_2a). For (m_3 = 5.00\ kg), (m_3g - T_2=m_3a). Adding these three equations: ((m_1 + m_2+m_3)g=(m_1 + m_2+m_3)a). The net - force equation for the whole system is ((m_3g+m_1g - m_2g)=(m_1 + m_2+m_3)a). Substitute (m_1 = 3.00\ kg), (m_2 = 4.00\ kg), (m_3 = 5.00\ kg) and (g = 9.8\ m/s^2). ((3 + 5-4)\times9.8=(3 + 4+5)a). (4\times9.8 = 12a).

Step2: Solve for acceleration (a)

[a=\frac{4\times9.8}{12}=\frac{9.8}{3}\approx3.27\ m/s^2]

Step3: Find tension (T_1)

From (m_1g - T_1=m_1a), we have (T_1=m_1(g - a)). Substitute (m_1 = 3\ kg), (g = 9.8\ m/s^2) and (a=\frac{9.8}{3}\ m/s^2). [T_1=3\times\left(9.8-\frac{9.8}{3}\right)=3\times\frac{2\times9.8}{3}=19.6\ N]

Step4: Find tension (T_2)

From (m_3g - T_2=m_3a), we have (T_2=m_3(g - a)). Substitute (m_3 = 5\ kg), (g = 9.8\ m/s^2) and (a=\frac{9.8}{3}\ m/s^2). [T_2=5\times\left(9.8-\frac{9.8}{3}\right)=5\times\frac{2\times9.8}{3}\approx32.7\ N]

Step5: Solve for acceleration in second problem

For the second problem, consider the two - mass system. The net - force acting on the system is the weight of the hanging mass. The total mass of the system is (M=20.0\ kg + 0.0150\ kg). The force causing the acceleration is (F = 0.0150\ kg\times g). Using (F = Ma), where (F=0.0150\times9.8\ N) and (M=(20.0 + 0.0150)\ kg). ((20.0 + 0.0150)a=0.0150\times9.8). [a=\frac{0.0150\times9.8}{20.0 + 0.0150}\approx0.00734\ m/s^2]

Answer:

  1. (a) The acceleration of each mass is approximately (3.27\ m/s^2). (b) The tension in the string between (m_1) and (m_2) is (19.6\ N) and the tension in the string between (m_2) and (m_3) is approximately (32.7\ N).
  2. The acceleration of the cart is approximately (0.00734\ m/s^2)